Probability Ques 81
- In a multiple-choice question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks the correct answers. The candidates decide to tick the answers at random, if he is allowed upto three chances to answer the questions, find the probability that he will get marks in the question.
$(1985,5$ M)
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Answer:
Correct Answer: 81.$\frac{1}{5}$
Solution:
Formula:
- The total number of ways to answer the question
$$ ={ }^{4} C _1+{ }^{4} C _2+{ }^{4} C _3+{ }^{4} C _4=2^{4}-1=15 $$
$P($ getting marks $)=P($ correct answer in I chance $)$
$+P$ (correct answer in II chance)
$+P($ correct answer in III chance)
$$ =\frac{1}{15}+\frac{14}{15} \cdot \frac{1}{14}+\frac{14}{15} \cdot \frac{13}{14} \cdot \frac{1}{13}=\frac{3}{15}=\frac{1}{5} $$
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